Question

Find the least number of six digits which is a perfect square. Find the square root of this number.

Answer 1

All 6 digit number starts from 1,00,000. 
Therefore the square of 'x' must be more then or equal to 100000

$x^{2} \geq 100000$

$x \geq \sqrt{100000}$

= 316.23
= 317 (round of to the larger value)

Now find the square of 317

$317^{2}$ = 100489

Answer 2

• Let us try 999999. The square root of 999999 = 999.9995, which is not a perfect square root.

• The nearest number is 999 and its square = 998001.

• The greatest 6-digit number that is a perfect square is 998001 and its square root is 999